Henderson-Hasselbalch Approximation - Saylor

Henderson-Hasselbalch Approximation

The Henderson-Hasselbalch formula allows us one method to approzimate the pH of a buffer
solution. The basic equation is as follows:

Motivation

We have straightforward caluclations for strong acids and bases, but the computations behind
buffers are rather complex and time consuming. By using the fact that weak acids and bases
barely ionize, allowing us to approximate the pH of buffer solutions using initial
concentrations. Though the approximation has a few restrictions, it simplifies a lenghty
calculation into a simple equation derived from K.

Brief History

Lawrence Joseph Henderson (1878-1942)

He was a talented biochemist, among many other titles, who spent most of his career at Harvard.

He was responsible for developing the components of the equation after studying equilibrium

reactions that took place within blood as a result of respiration (specializing in "fatigue"). His
equation was incomplete without a solid calculations going into it.M

Karl Albert Hasselbalch (1874-1962)

He was a chemist who studied pH closely. He also studied blood and reactions that took place

with oxygen, to put in the simplest of terms. He eventually modified Henderson's equation by
putting mathematical logs into it creating a solid relationship.J

Requirements

(1) -1 < log ( ) < 1

(2) The molarity of the buffer(s) should be 100x that of the acid ionization constant, or Ka.

(3) This equation will give poor or inaccurate results if there are strong acids or bases. pKa

values between 5 and 9 will give good approximations, but when we are out of this range there is
a strong chance that the pH value will be incorrect.J In other words near the end or the beginning

of a titration where the "relative concentrations of the acid or base differ substantially[,]...
approximate calculations break down."J

Derivation

Source URL: http://chemwiki.ucdavis.edu/Physical_Chemistry/Acids_and_Bases/Buffers/Henderson-Hasselbalch_Approximation
Saylor URL: http://www.saylor.org/courses/chem102/

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For a weak acid and its conjugate base :

which has an acid ionization constant
The Henderson-Hasselbalch equation is derived from this acid ionization constant.

Note that is a weak acid. By definition, does not dissociate completely and we can
and . Hence, we can use the initial concentrations because
say

Simillarly, for a weak base and its conjugate acid :

which has an base ionization constant
The Henderson-Hasselbalch equation is derived from this base ionization constant.

Note that is a weak base. By definition, does not dissociate completely and we can

say and . Hence, we can use the initial concentrations because

Source URL: http://chemwiki.ucdavis.edu/Physical_Chemistry/Acids_and_Bases/Buffers/Henderson-Hasselbalch_Approximation
Saylor URL: http://www.saylor.org/courses/chem102/

Attributed to: [wikipedia.org] www.saylor.org
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Examples

(1.) Find [H+] in a solution 1.0M HNO2 and 0.225M NaNO2 (Ka NO2 = 7.4*10-4)

(2.) What ratio will create an acetic acid buffer of pH 5.0? (Ka acetic acid = 1.8*10-5)
(3.) You prepare a buffer solution of .323 M NH3 and (NH4)2SO4. What molarity of (NH4)2SO4
is necessary to have a pH of 8.6? (pKb NH3= 4.74)
(4.) What is the pH of a buffer 0.500 moles acetic acid and 0.500 moles acetate ion and the total
volume is 5 L when you add 0.350 moles HCl? (Ka acetic acid = 1.8*10-5)
(5.) What is the range of an acetic acid buffer described in (4.) without the added HCl?

Solutions

(1.) pKa = -log Ka = -log(7.4*10-4) = 3.14

pH = pKa + log ( )

pH = 3.14 + log ( 1/0.225 )

pH = 3.14 + 0.648 = 3.788
[H+] = 10-pH = 10-3.788 = 1.6*10-4

(2.) pKa = -log Ka = -log(1.8*10-5) = 4.74

pH = pKa + log ( )

5.0 = 4.74 + log ( )

0.26 = log ( )
100.26 =

1.8 =

(3.) pKa + pKb = 14
pKa = 14 - 4.74 = 9.26

pH = pKa + log ( )

8.6 = 9.26 + log ( )

-0.66 = log ( )
[NH4+] = 1.48 M

(4.) pKa = -log Ka = -log(7.4*10-4) = 3.14

Source URL: http://chemwiki.ucdavis.edu/Physical_Chemistry/Acids_and_Bases/Buffers/Henderson-Hasselbalch_Approximation
Saylor URL: http://www.saylor.org/courses/chem102/

Attributed to: [wikipedia.org] www.saylor.org
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pH = pKa + log ( ) (note that mole ratio also works in place of concentrations

as both the acid and base are in the same solution)

pH = 3.14 + log (0.588)

pH = 3.14 - .230 = 3.37

(5.) pKa = -log Ka = -log(7.4*10-4) = 3.14
Note that the buffer range is of magnitude 2 pH units, with the pKa as the midpoint.
Hence, the buffer range would be 2.14 to 4.14.

References

1. J J. Chem. Educ. 2001, 78, 1499. (Journal Title - Henderson-Hasselbalch Equation: Its
History and Limitations)

2. D Madsen, Dorte. "Chapter 17." Chemistry 194, UC Davis, Davis. (Prof. Madsen's Ch. 17
Lecture Notes)

3. M National Academy of Sciences (U.S.) Biographical memoirs. City of Washington,
1945. Vol. XXIII, 2d memoir.

4. Chem. Petrucci, et al. General Chemistry: Principles & Modern Applications. 9th ed. Upper
Saddle River, New Jersey 2007.

Contributors

 Gurinder Khaira
 Alexander Kot (UCD)

Source URL: http://chemwiki.ucdavis.edu/Physical_Chemistry/Acids_and_Bases/Buffers/Henderson-Hasselbalch_Approximation
Saylor URL: http://www.saylor.org/courses/chem102/

Attributed to: [wikipedia.org] www.saylor.org
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